Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch) && ch != '-') ch = getchar();
  if (ch == '-') f = -1, ch = getchar();
  while (isdigit(ch)) x = (x << 1) + (x << 3) + ch - 48, ch = getchar();
  return x * f;
}
int n;
long long xL, xR, y, L[N], R[N], sum[N];
int main() {
  y = read(), xL = read(), xR = read();
  n = read();
  for (int i = 1; i <= n; i++)
    L[i] = read(), R[i] = read(), sum[i] = sum[i - 1] + R[i] - L[i];
  int q = read();
  while (q--) {
    long long px = read(), py = read();
    long double lv = (long double)px - (long double)(px - xL) / (py - y) * py;
    long double rv = (long double)px - (long double)(px - xR) / (py - y) * py;
    int a = 0, b = n + 1;
    for (int i = 20; i >= 0; i--)
      if (a + (1 << i) <= n && R[a + (1 << i)] <= lv) a += 1 << i;
    for (int i = 20; i >= 0; i--)
      if (b - (1 << i) >= 1 && L[b - (1 << i)] >= rv) b -= 1 << i;
    a++, b--;
    if (a == n + 1 || b == 0) {
      puts("0.0000000000");
      continue;
    }
    long double v = R[a] - max(lv, (long double)L[a]) +
                    min((long double)R[b], rv) - L[b] + sum[b - 1] -
                    sum[a + 1 - 1];
    printf("%.10lf\n", (double)(v * (py - y) / py));
  }
  return 0;
}